I am following a derivation in a book. It is implementing a state $|{\psi}\rangle$ into the eigenvalue equation $\hat{H}|{\psi}\rangle=E|{\psi}\rangle$. The $|{\psi}\rangle$ term contains a $|{\phi}\rangle-\Sigma_n|{\chi_n}\rangle...$ and when is expanded it becomes $\hat{H}|{\phi}\rangle-\Sigma_{n} E_n|{\chi_n}\rangle...$.

Could someone explain why the Hamiltonian becomes $E_n$ when acting within the sigma operator?

  • 1
    $\begingroup$ You might want to elaborate a bit more on the expanded line, but it looks like its just the eigenvalue of the n(th) state, i.e. the energy of the nth excited state of $\chi$ (look at the harmonic oscillator for example). This would mean that the sigma operator and the Hamiltonian commute. Do you have the form of the sigma operator? $\endgroup$
    – Akoben
    Dec 5 '14 at 11:22

It seems, the state $|\psi\rangle$ is a superposition of $|\phi\rangle$ and several eigenstates of the Hamiltonian:

$$ \hat{H}|\chi_n\rangle = E_n|\chi_n\rangle $$

The sigma then just denotes the sum of the eigenstates. And since the Hamilton operator is linear, you can easily apply it to each element of the sum independently.

$$ \hat{H}|\psi\rangle = \hat{H}\left(|\phi\rangle - \Sigma_n|\chi_n\rangle\right) = \hat{H}|\phi\rangle - \hat{H}\left(|\chi_1\rangle + |\chi_2\rangle + \dots \right) $$ $$ = \hat{H}|\phi\rangle - \left(\hat{H}|\chi_1\rangle + \hat{H}|\chi_2\rangle + \cdots\right) = \hat{H}|\phi\rangle - \left(E_1|\chi_1\rangle + E_2|\chi_2\rangle + \cdots\right) $$ $$ = \hat{H}|\phi\rangle - \Sigma_nE_n|\chi_n\rangle $$


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